### Exerccio 2 ###
	Melanie

#Biomassa de rvores#
b^ =e(-1.7953)*d^2.2974 #transformar "e" em exponencial: b^=(exp(-1.7953)*d^(2.2974))?)
ou
ln(b)^ = -2.6464+1.9960ln(d)+0.7558ln(h) #ln(b)^ = exp(-2.6464 + 1.9960*log(15) + 0.7558*log(1200))

DAP=15cm
h=12m=1200cm

1a frmula: b=83.61095
2a frmula: ln(b)=103.2301

Resposta: Sim, os modelos resultaro em estimativas distintas.

#Sequncias#
1: a a a a a a 
a=c(rep("a", times=6))
2: 1 1 1 2 2 2 3 3 3
b=c(rep(1:3, each=3))
3: 1 1 1 2 2 3
c=c(rep(1,3),rep(2,2),rep(3,1))
4: 1 2 3 4 5 4 3 2 1
d=c(seq(1,5),seq(4,1))
5: Nmeros mpares de 1 a 99
e=c(seq(1,99, by=2))

#Conta de luz#
1.
leituras=c(9839, 10149, 10486, 10746, 11264, 11684, 12082, 12599, 13004, 13350, 13717, 14052)
consumo.mensal=diff(leituras)= 310 337 260 518 420 398 517 405 346 367 335
2.
max(consumo.mensal)
[1] 518
min(consumo.mensal)
[1] 260
range(consumo.mensal)
[1] 260 518
3.
mean(consumo.mensal)
[1] 383
median(consumo.mensal)
[1] 367
var(consumo.mensal)
[1] 6476.2

#rea Basal#
1.
d=13.5
A=pi*(d/2)^2
2.
d1=7
d2=9
d3=12
A2=(pi*(d1/2)^2)+(pi*(d2/2)^2)+(pi*(d3/2)^2)

#Varincia na Unha#
1.
pesos <- c(78.4, 79.8, 76.0, 75.3, 77.4, 78.6, 77.9, 78.8, 79.2, 75.2, 75.0, 79.4)
a=mean(pesos)
b=length(pesos)
var=sum((a-pesos)^2)/(b-1)
var=3.110606

dp=sqrt(var)
dp=1.763691

2.
var(pesos)
[1] 3.110606
sd(pesos)
[1] 1.763691

#Teste t de Student#
?rt
pt(q, df, ncp, lower.tail = FALSE)
a=pt(2.2,19,lower.tail=FALSE)*2
a
[1] 0.0403811, <0.05 portanto significativo
b=pt(1.9,19,lower.tail=FALSE)*2
b
[1] 0.0727184, >0.05 portanto no significativo

